Question

Find the sum of each pair of vectors and match it with the magnitude of the resultant vector. PLEASE HELP. Information on the picture

Answer 1

Answer: 1. 5.83 m/s > magnitude 4.5 m/s, direction angle 55˚ magnitude 3

m/s, direction angle 135°

2. 4.05 m/s > magnitude 3.5 m/s, direction angle 35˚

magnitude 4 m/s, direction angle 150˚

3. 5.29 m/s > magnitude 6 m/s, direction angle 120˚

magnitude 2 m/s, direction angle 240˚

4. 3.32 m/s > magnitude 3 m/s, direction angle 70˚

magnitude 5 m/s, direction angle 210˚

Refer to picture below.

Step-by-step explanation: I got this correct on Edmentum.

Answer 2

1. Magnitude 3.5 m/s, direction angle 35°, magnitude 4 m/s, direction angle 150°.

Matches to 4.05 m/s

First of all, let's write this statement in vector form. For the fist vector we have:

Magnitude 3.5 m/s, direction angle 35°:

Let's say this is vector
`\vec{A}`, so the magnitude is:

`\left|\vec{A}\right|=3.5m/s`

And the direction is defined as:

`\theta = 35^(\circ)`

So the components are:

`Ax=\left|\vec{A}\right| cos\theta \\ \\ Ax=3.5 cos35^(\circ)=2.86m/s \\ \\ \\ Ay=\left|\vec{A}\right| sin\theta \\ \\ Ay=3.5 sin35^(\circ)=2m/s`

So vector
`\vec{A}` is:

`\vec{A}=2.86i+2j`

For the second vector:

Magnitude 4 m/s, direction angle 150°:

Let's say this is vector
`\vec{B}`, so the magnitude is:

`\left|\vec{B}\right|=4m/s`

And the direction is defined as:

`\theta = 150^(\circ)`

So the components are:

`Bx=\left|\vec{B}\right| cos\theta \\ \\ Bx=4 cos150^(\circ)=-2√(3)m/s \\ \\ \\ By=\left|\vec{B}\right| sin\theta \\ \\ By=4 sin150^(\circ)=2m/s`

So vector
`\vec{B}` is:

`\vec{B}=-2√(3)i+2j`

THE SUM OF THESE TWO VECTORS IS:

`\vec{R}=\vec{A}+\vec{B}=(2.86i+2j)+(-2√(3)i+2j) \\ \\ \boxed{\vec{R}=-0.60i+4j}`

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

`\left|\vec{R}\right|=√(Rx^2+Ry^2) \\ \\ Rx=Ax+Bx \\ \\ Ry=Ay+By \\ \\ \\ \left|\vec{R}\right|=√((-0.6^2)+(4)^2) \\ \\ \boxed{\left|\vec{R}\right|=4.05m/s}`

2. Magnitude 4.5 m/s, direction angle 55°, magnitude 3 m/s, direction angle 135°.

Matches to 5.83 m/s

Magnitude 4.5 m/s, direction angle 55°:

Let's say this is vector
`\vec{C}`, so the magnitude is:

`\left|\vec{C}\right|=4.5m/s`

And the direction is defined as:

`\theta = 55^(\circ)`

So the components are:

`Cx=\left|\vec{C}\right| cos\theta \\ \\ Cx=4.5 cos55^(\circ)=2.58m/s \\ \\ \\ Cy=\left|\vec{C}\right| sin\theta \\ \\ Cy=4.5 sin55^(\circ)=3.68m/s`

So vector
`\vec{C}` is:

`\vec{C}=2.58i+3.68j`

For the second vector:

Magnitude 3 m/s, direction angle 135°:

Let's say this is vector
`\vec{D}`, so the magnitude is:

`\left|\vec{D}\right|=3m/s`

And the direction is defined as:

`\theta = 135^(\circ)`

So the components are:

`Dx=\left|\vec{D}\right| cos\theta \\ \\ Dx=3 cos135^(\circ)=-(3√(2))/(2) \\ \\ \\ Dy=\left|\vec{D}\right| sin\theta \\ \\ Dy=3 sin135^(\circ)=(3√(2))/(2)`

So vector
`\vec{D}` is:

`\vec{D}=-(3√(2))/(2)i+(3√(2))/(2)j`

THE SUM OF THESE TWO VECTORS IS:

`\vec{R}=\vec{C}+\vec{D}=(2.58i+3.68j)+(-(3√(2))/(2)i+(3√(2))/(2)j) \\ \\ \boxed{\vec{R}=0.46i+5.80j}`

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

`\left|\vec{R}\right|=√((0.46)^2+(5.8)^2) \\ \\ \boxed{\left|\vec{R}\right|=5.83m/s}`

3. Magnitude 3 m/s, direction angle 70°, magnitude 3 m/s, direction angle 135°.

Matches to 3.32 m/s

Magnitude 4.5 m/s, direction angle 55°:

This is vector
`\vec{E}`, so the magnitude is:

`\left|\vec{E}\right|=3m/s`

Direction:

`\theta = 70{\circ}`

Components:

`Ex=\left|\vec{E}\right| cos\theta \\ \\ Ex=3 cos70^(\circ)=1.02m/s \\ \\ \\ Ey=\left|\vec{E}\right| sin\theta \\ \\ Ey=3 sin70^(\circ)=2.82m/s`

So:

`\vec{E}=1.02i+2.82j`

For the second vector:

Magnitude 5 m/s, direction angle 210°:

`\vec{F}`:

`\left|\vec{F}\right|=5m/s`

Direction:

`\theta = 210^(\circ)`

Components:

`Fx=5 cos210^(\circ)=-(5√(3))/(2) \\ \\ \\ Ey=5 sin210^(\circ)=-(5)/(2)`

Then:

`\vec{F}=-(5√(3))/(2) i-(5)/(2)j`

THE SUM OF THESE TWO VECTORS IS:

`\vec{R}=(1.02i+2.82j)+(-(5√(3))/(2)i-(5)/(2)j) \\ \\ \boxed{\vec{R}=-3.31i+0.32j}`

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

`\left|\vec{R}\right|=√((-3.31)^2+(0.32)^2) \\ \\ \boxed{\left|\vec{R}\right|=3.32m/s}`

4. Magnitude 6 m/s, direction angle 120°, magnitude 2 m/s, direction angle 140°.

Matches to 5.29 m/s

Magnitude 6 m/s, direction angle 120°:

`\left|\vec{W}\right|=6m/s`

Direction:

`\theta = 120^(\circ)`

Components:

`Wx=6 cos120^(\circ)=-3m/s \\ \\ \\ Wy=6 sin120^(\circ)=3√(3)m/s`

So:

`\vec{W}=-3i+3√(3)j`

For the second vector:

Magnitude 2 m/s, direction angle 240°:

`\vec{Z}`:

`\left|\vec{Z}\right|=2m/s`

Direction:

`\theta = 240^(\circ)`

Components:

`Zx=2 cos240^(\circ)=-1 \\ \\ \\ Zy=2 sin240^(\circ)=-√(3)`

Then:

`\vec{Z}=-i-√(3)j`

THE SUM OF THESE TWO VECTORS IS:

`\vec{R}=(-3i+3√(3)j)+(-i-√(3)j) \\ \\ \boxed{\vec{R}=-4i+2√(3)j}`

THE MAGNITUDE OF THE RESULTANT VECTOR IS:

`\left|\vec{R}\right|=\sqrt{(-4)^2+(2√(3))^2} \\ \\ \boxed{\left|\vec{R}\right|=5.29m/s}`